chapter 14: phase equilibria applications part ii
TRANSCRIPT
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Chapter 14: Phase Equilibria Applications
Part II
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If the component is supercritical, then the vapor pressure is not defined
111
101
ˆlim
1f
x
fH x
P
Pxy
P
HxyPyHx
Hxf
sat
l
2
2222
1
11
11
11111
11
11
111
ˆˆ
ˆ
2
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example• A binary methane (1) and a light oil (2) at 200K and 30 bar
consists of a vapor phase containing 95% methane and a liquid phase containing oil and dissolved methane. The fugacity of methane is given by Henry’s law, and at 200 K, H1 = 200 bar. Estimate the equilibrium mole fraction of methane in the liquid phase. The second virial coefficient of methane at 200K is -105 cm3/mol
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solution
:
ˆˆ
ˆ
111
111
sAssumption
Pyf
Hxfv
l
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Solution (cont)
PyPyf
Hxfv
l
11111
111
ˆˆ
ˆ
Need equation for the fugacity coefficient vapor phase
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Solution (cont)
827.0)/exp( 111 RTPB
How do we solve for the mole fraction of the solutein the liquid phase?
x1 =0.118
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Another example
• For chloroform(1)/ethanol(2) at 55oC, the excess Gibbs energy is
2121 )59.042.1( xxxxRT
G E
• The vapor pressures of chloroform and ethanol at 55oC are P1
sat = 82.37 kPa, and P2sat = 37.31 kPa
• Make BUBLP calculations, knowing that B11=-963 cm3/mol, B22 =-1,523 cm3/mol, B12 =52 cm3/mol
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)1)((2()(exp
)(2()1(exp
12112212
12
11221122
11
xAAAx
xAAAx
satsat2
22
1
11
ˆ;
ˆ
Need equations for the fugacity coefficients
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2
satsat
and calculate
BBB
RT
PyPB
RT
PB
1
22111212
122211
1
1111
2
expˆ
exp
But we don’t know P
Guess P (avg sat. pressures)
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12
21222122
11112111
1
)(),,,(
)(),,,(
yy
PxxPyyTPy
PxxPyyTPysat
sat
For example, at x1 =0.25, solve for y1, y2, P
y1 = 0.558y2 = 0.442P = 63.757 kPa
In our web site, there are model spreadsheets that you can download 10
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VLE from cubic EOS
PxPy
NDESCRIPTIO EALTERNATIV
...N 2, 1, i ff
lii
vii
li
vi
ˆˆ
,ˆˆ
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Vapor pressure pure species
Liquid branch
Vapor branch
LV transition
li
vi
Using equation (1), the cubic EOS yields Pisat= f(T)
(1)
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Compressibility factors
ii
lii
ilii
lii
li q
ZZZZ
1
))((1
ci
cirii
i
ii
ci
cii
ii
P
TRTTa
RTb
Taq
P
RTb
RT
Pb
22)()(
)(
For the vapor phase there is anotherexpression, (14.36)
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How to calculate vapor pressure from a cubic EOS
ili
ilil
i
iiiiii
Z
ZI
where
IqZZ
ln1
)ln(1ln
We solve for the saturation pressure at a given T such that the fugacity coefficients are equal in the two phases: 2 compressibility eqns., two fugacity coefficient eqns., equality of fugacity coefficients, 5 unknowns: Psat, Zl, Zv, l, v
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Mixture VLE from a cubic EOS
• Equations for Zl and Zv have the same form as for pure components
• However the parameters a & b are functions of composition
• The two phases have different compositions, therefore we could think in terms of two P-V isotherms, one for each composition
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Mixing rules for parameters
2/1jijiij
iji j
ji
ii
i
aaaa
axxa
bxb
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Also we need “partial” parameters
j
j
j
nTii
nTii
nTii
n
nqq
n
nbb
n
naa
,
,
,
)(
)(
)(
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The partial parameters are used for the calculation of fugacity
coefficients• Because i is related to the partial molar
property of GR (residual G)
b
b
a
aqq
I
IqZZb
b
iii
ii
i
1
different) are (equations or Z Zis Z
17) (slide before as defined
)ln()1(ˆln
vl
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example
• Vapor mixture of N2(1) and CH4 (2) at 200K and 30 bar contains 40 mol% N2. Calculate the fugacity coefficients of nitrogen and methane using the RK equation of state.
vvvv
vvvvvv
ZZ
ZqZ
1
For RK, =0 and =1
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molcmP
Tb
molcm barP
TTa
b
b
a
aayayqq
aaayaya
ayaayyaya
ci
cii
6
ci
cirii
/14.83
08664.0
/)14.83(
42748.0
22
22
2
3
2222/1
1212111
212111
22221211
21
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Calculate P-x-y diagram at 100 oF for methane(1)-n butane (2) using SRK and mixing rules (14.42) to
(14.44)
Compare with published experimental data (P, x, y)
Initial values for P and yi can be taken from given experimental data
First read critical constants, , from Table B.1 and from Table 3.1
22/12 )1)(176.0574.148.0(1 rSRK T
Calculate b1, b2, a1, a2
In this case T > Tc1 26
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K value given by
vi
li
iKˆ
ˆ
Step 1
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The equations for are valid only up to the critical temperature; however is OK toextend the correlation slightly above the critical temperature
Lets calculate the mixture parameters (for step 1). When applied to the liquid phase we use the xi mole fractions
RTb
aq
b
b
a
aaxaxqq
b
b
a
aaxaxqq
bxbxb
axaaxxaxa
l
ll
llll
llll
l
l
2211222
1212111
2211
22221211
21
22
22
2
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Follow diagram Fig. 14.9
Assume P and yi
Calculate Zl and Zv, and the mixture fugacity coefficients
Calculate K1 and K2 and the Kixi
Calculate normalized yi=Kixi/ Kixi
Reevaluate fugacity coefficients vapor phase, etc
If Kixi > 1, P is too low so increase P; if Kixi < 1, then reduce P
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Results:Rms % difference between calculated and exp. P is3.9%
Rms deviation between calculated and exp. y1 is0.013
Note that the system consistsof two similar molecules
Where are the largestdiscrepancies with the experimental data?
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